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2 years ago

This is brilliant!!! I hope this helps yall get out of studying Lol

Mathematics

2 answers:

qaws [65]2 years ago

8 0

Thank you so much lol

givi [52]2 years ago

4 0

Answer:

lol

Step-by-step explanation:

I am going to show this to my teacher tomorrow

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A box-and-whisker plot. The number line goes from 0 to 9. The whiskers range from 0 to 7, indicated by A and E. The box ranges f

Sladkaya [172]

minimum value: 1

lower quartile: 3

median of the data: 6

upper quartile: 8

maximum value: 14

Sorry for the late answer, i hope this will help you in the future and everyone else who reads this.

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2 years ago

A store buys 17 sweaters for $255 and sells them for $578. How much profit 1 point

ddd [48]

The answer is 243 is the answer to this question

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2 years ago

I need help this is due tomorrow !

natulia [17]

Answer:

b i think

Step-by-step explanation:

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2 years ago

Read 2 more answers

_________ integers are natural numbers. 1.) some 2.) all 3.) no

Marina CMI [18]

Answer:

1.) Some

Step-by-step explanation:

Natural numbers do not contain negative numbers, but they do contain whole numbers like 1, 2, 3. So your answer would be some.

6 0

3 years ago

Let f(x) = [infinity] xn n2 n = 1. find the intervals of convergence for f. (enter your answers using interval notation. ) find

inna [77]

Best guess for the function is

$\displaystyle f(x) = \sum_{n=1}^\infty \frac{x^n}{n^2}$

By the ratio test, the series converges for

$\displaystyle \lim_{n\to\infty} \left|\frac{x^{n+1}}{(n+1)^2} \cdot \frac{n^2}{x^n}\right| = |x| \lim_{n\to\infty} \frac{n^2}{(n+1)^2} = |x| < 1$

When $x=1$ , $f(x)$ is a convergent $p$ -series.

When $x=-1$ , $f(x)$ is a convergent alternating series.

So, the interval of convergence for $f(x)$ is the closed interval $\boxed{-1 \le x \le 1}$ .

The derivative of $f$ is the series

$\displaystyle f'(x) = \sum_{n=1}^\infty \frac{nx^{n-1}}{n^2} = \frac1x \sum_{n=1}^\infty \frac{x^n}n$

which also converges for $|x|$ by the ratio test:

$\displaystyle \lim_{n\to\infty} \left|\frac{x^{n+1}}{n+1} \cdot \frac n{x^n}\right| = |x| \lim_{n\to\infty} \frac{n}{n+1} = |x| < 1$

When $x=1$ , $f'(x)$ becomes the divergent harmonic series.

When $x=-1$ , $f'(x)$ is a convergent alternating series.

The interval of convergence for $f'(x)$ is then the closed-open interval $\boxed{-1 \le x < 1}$ .

Differentiating $f$ once more gives the series

$\displaystyle f''(x) = \sum_{n=1}^\infty \frac{n(n-1)x^{n-2}}{n^2} = \frac1{x^2} \sum_{n=1}^\infty \frac{(n-1)x^n}{n} = \frac1{x^2} \left(\sum_{n=1}^\infty x^n - \sum_{n=1}^\infty \frac{x^n}n\right)$

The first series is geometric and converges for $|x|$ , endpoints not included.

The second series is $f'(x)$ , which we know converges for $-1\le x$ .

Putting these intervals together, we see that $f''(x)$ converges only on the open interval $\boxed{-1 < x < 1}$ .

6 0

1 year ago